UNIT-5

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UNIT-5
One of the endlessly alluring aspects of mathematics is that its thorniest
paradoxes have a way of blooming into beautiful theories
1. The fourth term of an AP is 0. Prove that its 25th term is triple its 11th term.
Ans:
a4= 0
a + 3d = 0
T.P
a25= 3 (a11)
a + 24d = 3 (a + 10d)
a + 24d = 3a + 30d
RHS sub a = - 3d
- 3d + 24d = 21d
LHS 3a + 30d
- 9d + 30d = 21d
LHS = RHS
Hence proved
2. Find the 20th term from the end of the AP 3, 8, 13……..253.
Ans: 3, 8, 13 ………….. 253
Last term = 253
a20 from end
= l – (n-1)d
253 – ( 20-1) 5
253 – 95
= 158
3. If the pth, qth & rth term of an AP is x, y and z respectively,
show that x(q-r) + y(r-p) + z(p-q) = 0
Ans: pth term x = A + (p-1) D
qth term y = A + (q-1) D
rth term z = A + (r-1) D
T.P x(q-r) + y(r-p) + z(p-q) = 0
={A+(p-1)D}(q-r) + {A + (q-1)D} (r-p)
+ {A+(r-1)D} (p-q)
A {(q-r) + (r-p) + (p-q)} + D {(p-1)(q-r)
+ (r-1) (r-p) + (r-1) (p-q)}
A.0 + D{p(q-r) + q(r-p) + r (p-q)
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- (q-r) – (r-p)-(p-q)}
= A.0 + D.0 = 0.
Hence proved
4. Find the sum of first 40 positive integers divisible by 6 also find the sum of first 20 positive
integers divisible by 5 or 6.
No’s which are divisible by 6 are
6, 12 ……………. 240.
Ans:
40
S40 =
6
240
2
= 20 x 246
= 4920
No’s div by 5 or 6
30, 60 …………. 600
20
30
600
2
= 10 x 630
= 6300
5. A man arranges to pay a debt of Rs.3600 in 40 monthly instalments which are in a AP.
When 30 instalments are paid he dies leaving one third of the debt unpaid. Find the value
of the first instalment.
Ans: Let the value of I instalment be x
40
2a
39 d
2
=3600
2a + 39d = 180
S30 =
30
2a
29 d
2
S40 = 3600.
-
1
-
2
=2400
30a + 435d = 2400
2a + 29d = 160
Solve 1 & 2 to get
d = 2 a = 51.
I instalment = Rs.51.
6. Find the sum of all 3 digit numbers which leave remainder 3 when divided by 5.
Ans: 103, 108……….998
a + (n-1)d
=
998
103 + (n-1)5 =
998
n
=
180
S180
=
180
103
998
2
= 90 x 1101
S180
=
99090
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7. Find the value of x if 2x + 1, x2 + x +1, 3x2 - 3x +3 are consecutive terms of an AP.
Ans:
a2 –a1 = a3 –a2
x2 + x + 1-2x - 1 = 3x2 – 3x + 3- x2-x-1
x2 - x = 2x2 – 4x + 2
x2 - 3x + 2 = 0
(x -1) (x – 2) = 0
x = 1 or x = 2
8. Raghav buys a shop for Rs.1,20,000.He pays half the balance of the amount in cash and
agrees to pay the balance in 12 annual instalments of Rs.5000 each. If the rate of interest is
12% and he pays with the instalment the interest due for the unpaid amount. Find the total
cost of the shop.
Ans: Balance = Rs.60,000 in 12 instalment of Rs.5000 each.
12
Amount of I instalment
= 5000 +
II instalment
= 5000 + (Interest on unpaid amount)
100
60,000
= 5000 + 6600
12
x 55000
100
= 11600
III instalment
= 5000 + (Interest on unpaid amount of Rs.50,000)
AP is 12200, 11600, 11000
D = is 600
Cost of shop = 60000 + [sum of 12 instalment]
= 60,000 +
12
2
[24,400-6600]
= 1,66,800
9. Prove that am + n + am - n =2am
Ans: a m + n = a1 + (m + n - 1) d
a m-n = a1 + (m - n -1) d
am = a1 + (m-1) d
Add 1 & 2
a m+n + a m-n =
a1+(m+n-1) d+ a1 + (m-n-1)d
=
2a1+(m+n+m-n-1-1)d
=
2a1+ 2(m-1)d
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=
=
=
2[a1+ (m-1)d]
2[a1+ (m-1)d]
2am. Hence proved.
10. If the roots of the equation (b-c)x2 +(c-a)x +(a-b) = 0 are equal show that a, b, c are in AP.
Ans: Refer sum No.12 of Q.E.
If (b-c)x2 + (c-a)x + (a-b)x have equal root.
B2-4AC=0.
Proceed as in sum No.13 of Q.E to get c + a = 2b
b-a=c-b
a, b, c are in AP
11. Balls are arranged in rows to form an equilateral triangle .The first row consists of one ball,
the second two balls and so on. If 669 more balls are added, then all the balls can be
arranged in the shape of a square and each of its sides then contains 8 balls less than each
side of the triangle. find the initial number of balls.
Ans: Let their be n balls in each side of the triangle
No. of ball (in ) = 1 + 2+ 3………..=
n n
1
2
No. of balls in each side square = n-8
No. of balls in square = (n-8)2
APQ
n n
1
2
+ 660 = (n-8)2
On solving
n2 + n + 1320 = 2(n2 - 16n + 64)
n2 - 33n - 1210 = 0
(n-55) (n+22) = 0
n=-22 (N.P)
n=55
No. of balls =
n n
1
2
=
55 x 56
2
= 1540
12. Find the sum of (1
Ans:
1
1
n
1
2
n
1
)
2
(1
n
)
n
(1
3
n
) …….
upto n terms.
- upto n terms
1
2
n
n
[1+1+…….+n terms] – [ +
+….+ n terms]
n –[Sn up to n terms]
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n
Sn = [2a + (n-1)d]
2
=
=
n
2
2
n
n
1
n
2
n
, a=
1
n
)
1
(on simplifying)
1
2
n 1
1)
1
n
2
n=
(n
(d =
=
Ans
13. If the following terms form a AP. Find the common difference & write the next 3 terms3,
3+ 2, 3+2 2, 3+3 2……….
Ans: d=
2 next three terms 3 + 4
2
2
, 3+5
, 3+6
2
……..
14. Find the sum of a+b, a-b, a-3b, …… to 22 terms.
Ans: a + b, a – b, a – 3b, up to 22 terms
d= a – b – a – b = 2b
S22 =
22
2
[2(a+b)+21(-2b)]
11[2a + 2b – 42b]
= 22a – 440b Ans.
15. Write the next two terms 12, 27, 48, 75……………….
Ans: next two terms 108 , 147 AP is 2 3 , 3 3 , 4 3 , 5 3 , 6 3 , 7 3 ……
16. If the pth term of an AP is q and the qth term is p. P.T its nth term is (p+q-n).
Ans: APQ
ap = q
aq = p
an = ?
a + (p-1) d = q
a + (q-1) d = p
d[p – q] = q – p
Sub d = -1 to get
an = a + (n – 1)d
= a + (n - 1)d
= (q + p – 1) + (n – 1) - 1
an = (q + p – n)
= -1
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a = q + p -1
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17. If
Ans:
1
x
1
1
,
2 x
1
,
2 x
x
x
3
x
x
1
2
3 x
1
,
3 x
1
1
1
,
2
5
are in AP find x.
are in AP find x.
5
1
x
1
5
x
3
8x
15
2
5x
6
x
2
On solving we get x = 1
18. Find the middle term of the AP 1, 8, 15….505.
Ans: Middle terms
a + (n-1)d = 505
a + (n-1)7 = 505
n–1=
504
7
n = 73
37th term is middle term
a37 = a + 36d
= 1 + 36(7)
= 1 + 252
= 253
19. Find the common difference of an AP whose first term is 100 and sum of whose first 6
terms is 5 times the sum of next 6 terms.
Ans: a = 100
APQ a1 + a2 + ……. a6 =5 (a7 + …….. + a12)
a1
a6
6
2
=5 x 6
a7
a 12
2
a + a + 5d = 5[a + 6d + a + 11d]
8a + 80d = 0 (a = 100)
d = - 10.
20. Find the sum of all natural no. between 101 & 304 which are divisible by 3 or 5.
Find their sum.
Ans: No let 101 and 304, which are divisible by 3.
102, 105………….303 (68 terms)
No. which are divisible by 5 are 105, 110……300 (40 terms)
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No. which are divisible by 15 (3 & 5) 105, 120…… (14 terms)
There are 94 terms between 101 & 304 divisible by 3 or 5. (68 + 40 – 14)
S68 + S40 – S14
= 19035
21. The ratio of the sum of first n terms of two AP’s is 7n+1:4n+27. Find the ratio of their
11th terms .
Ans: Let a1, a2… and d1, d2 be the I terms are Cd’s of two AP’s.
Sn of one AP
7n
=
1
4n
27
Sn of II AP
m
2
m
2 a1
2
(n
2a2
(n
2 a1
(n
1) d 1
2a2
(n
1) d 2
1) d 1
7n
4n
1) d 2
7n
4n
1
27
1
27
We have sub. n = 21.
2 a1
20 d 1
2a2
20 d 2
=
7 x 21
4 ( 21 )
a1
10 d 1
148
a2
10 d 2
111
1
27
4
3
ratio of their 11th terms = 4 :3.
22. If there are (2n+1)terms in an AP ,prove that the ratio of the sum of odd terms and the sum
of even terms is (n+1):n
Ans: Let a, d be the I term & Cd of the AP.
ak = a + (k – 1) d
s1 = sum to odd terms
s1 = a1 + a3 + ……… a 2n + 1
s1 =
=
n
1
2
1
n
2
a1
a 2n
2a 1
2 nd
1
s1 = (n + 1) (a + nd)
s2 = sum to even terms
s2 = a2 + a4 + ….. a 2n
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s2 =
n
a2
2
a 2n
n
= [a + d + a + (2n – 1)d]
2
=n [a + nd]
s1 : s2 =
=
n
(n
1)( a
n(a
nd )
nd )
1
n
23. Find the sum of all natural numbers amongst first one thousand numbers which are neither
divisible 2 or by 5
Ans: Sum of all natural numbers in first 1000 integers which are not divisible by 2 i.e. sum of
odd integers.
1 + 3 + 5 + ………. + 999
n = 500
S500 =
500
2
[1 + 999]
= 2,50,000
No’s which are divisible by 5
5 + 15 + 25 …….. + 995
n = 100
Sn =
100
2
[5 + 995]
= 50 x 1000 = 50000
Required sum = 250000 – 50,000
= 200000
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